Q6 / 77 · NTA Abhyas (JEE Main) 2020 · Mathematics · Statistics

The mean deviation of the series $a^{2}, a^{2} + d, a^{2} + 2 d, . . . . ., a^{2} + 2 n d$ from its median is

A
$\frac{(n + 1) d}{(2 n + 1)}$
B
$\frac{n d}{2 n + 1}$
Correct answer
C
$\frac{n (n + 1) d}{(2 n + 1)}$
D
$\frac{(2 n + 1) d}{n (n + 1)}$

Solution

Median or mean

(\bar{x}) = \frac{\frac{2 n + 1}{2} \cdot (a^{2} + a^{2} + 2 n d)}{(2 n + 1)}

= a^{2} + n d

\{∵ M e a n = M e d i a n\}

Σ |x_{i} - \bar{x}| = 2 d (1 + 2 + . . . . + n) = n (n + 1) d

∴ M . D . = \frac{n (n + 1) d}{2 n + 1}

Hence, the correct answer is option: 'C'.

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